Check your email regarding the Practice Test Session.
For the Homework Buddy system, you're suppose to put A2D (agree to disagree) in the situation where you've discussed the discrepancy and somehow feel your answer is right and the other person won't listen to you. Otherwise, you should typically convince the other person why your answer is right and get them to change it. Future checks of the use of A2D will be more strict. And in general, listing another person as your Homework Buddy should mean ALL your answers match up. Perhaps not the work, but at least the answers. For now, I will also reduce the Homework Buddy bonus from 5 to 4. Until better use of the system occurs
Someone came to ask about the difference between a converging sequence and series.
ReplyDeleteThe most illuminating piece of information I gave them was that the series $$\sum_{n=1}^\infty a_n =\lim _{k\to \infty} b_k$$ where $$b_k=\sum_{n=1}^k a_n$$.
Thus, while we can have $$\lim_{n\to \infty} a_n =0$$, we might not have $$\lim _{k\to\infty} b_k <\infty$$
Let's derive the formula for a geometric series.
ReplyDeleteAt first, let
$$S_n = a + a\cdot r + \cdots +a \cdot r^n$$
Then
$$S_n r = a\cdot r + \cdots +a \cdot r^n+ a\cdot r^{n+1}$$
Then
$$S_n-S_n r=a-a\cdot r^{n+1}$$
Then
$$S_n=\frac{a-a \cdot r^{n+1}}{1-r}$$
If $\left| r\right|<1$, then $r^{n+1}$ goes to $0$ as $n$ goes to $\infty$.
Then
$$\sum_{n=0}^\infty a\cdot r^n=\lim_{k\to \infty} S_k =\frac{a}{1-r}$$
Does the Limit Comparison Test work exactly the same for alternating series? If there is a (-1)^n term does it change the convergence or divergence of the series?
ReplyDeleteIn the example that you've given, the equation goes back and forth between -1 and 1, without ever closing the gap between the two, and thus does not converge to a single point. I think this should be true for alternating series, but I think it may be a case by case basis when the alternating series is multiplied by another term.
DeleteI think Josh is right because if you have an alternating series with n! in the denominator, the numbers get smaller and smaller so it converges.
DeleteOkay thank you!
DeleteLove the brainstorming.
DeleteShort answer (to first question): no.
Limit Comparison Test:
Suppose that $\sum a_n$ and $\sum b_n$ are series with positive terms. If $$lim_{n\to\infty} \frac{a_n}{b_n}=c$$ where $c$ is a finite number and $c>0$, then either both series converge or both diverge.
If we don't assume that all the terms are positive, we can create counter-examples.
Counter-Example:
$$\sum (-1)^n \frac{1}{n}$$ converges, $$\sum (-1)^n\frac{1}{n}\left( 1+(-1)^n\frac{1}{\log(n)} \right)$$ diverges, and $$\lim_{n\to \infty} \frac{a_n}{b_n}=1$$.
Though Process Behind Counter-Example:
Looking at the proof (on page 724), it's clear why one would like the terms to all be positive. Though we might ask ourselves, can we generalize the conditions?
At first we have two series $\sum a_n $ and $\sum b_n$. Now assume $$\lim_{n\to \infty} \frac{a_n}{b_n} =c$$ where $0 N$, $$m < \frac{a_n}{b_n} < M$$. This means that $a_n$ and $b_n$ for $n > N$ have the same sign.
Staring at this fact long enough, a picture starts to form in my mind. Reason that if we had a series which converged, but had infinitely many positive and negative terms, then perhaps the negative terms are helping cancel the growth of the positive terms or vice versa. Given the converging series, we want to make a new series from the first, by slightly increasing the positive terms, and slightly increasing (less negative) the negative terms. Then we hope that this new series, as a result, diverges. For every negative number $b_n$, we want the $a_n$ to not subtract as much, so $a_n=b_n(C-\varepsilon_n)$ and for positive $b_n$ we want $a_n$ to be larger, and contribute more, so $a_n=b_n (C+\varepsilon_n)$. We introduce the sign function: $\mathrm{sgn}(b_n)=1$ for $b_n>0$ and $\mathrm{sgn}(b_n)=-1$ for $b_n < 0$. This allows us to write $a_n = b_n (C + \mathrm{sgn}(b_n)\varepsilon _n )$. Note that the absolute value of $b_n$ is $\mathrm{sgn}(b_n) \cdot b_n$. Then we have $$\sum a_n = C\sum b_n + \sum b_n \mathrm{sgn} (b_n) \varepsilon_n = \sum \left| b_n \right| \varepsilon_n$$.
We use the above paragraph and try to alter our favorite alternating series which doesn't converge when we take the absolute value of it's terms: $$\sum (-1)^n \frac{1}{n}$$. We want $$\sum \frac{1}{n} \varepsilon_n$$ to diverge, but for $$\lim_{n\to \infty} \varepsilon_n=0$$. This leads us to the above counter-example.
April 5, 2012 7:34 PM
Remark:
DeleteWe can apply the test if eventually the terms of each series has the same sign.
Example:
We can compare $\sum -\frac{1}{n}$ and $\sum \frac{2}{n+3}$. Technically we are applying the test to $\sum \frac{1}{n}$ and $\sum \frac{2}{n+3}$. We can compare $\sum_{n=1}^\infty \frac{1}{n}$ and $1-\frac{1}{2}+2-3+\sum_{n=6}^\infty \frac{1}{n}$.
For your second question, consider the Alternating Series Test on page 727.
Thank you for such a thorough response Tim!
DeleteYea, this is cool, thanks Tim.
DeleteTim,
ReplyDeleteIs it possible for a series to be alternating between negative and positives and geometric? I think it is, but I wanna make sure.
Please read the box on page 706 and give me the answer.
DeleteI'm confused as to how to figure out if a series converges or diverges when the variable is in the exponent. For example 11.4.20: sum from 1 to infinity of:
ReplyDelete(n + 4^n)/(n + 6^n).
Could you do some similar problems in section? Maybe 11.4.15?
Sure, I'll try to remember, otherwise remind me.
DeleteFrom an intuitive standpoint, the dominant terms in the numerator and denominator are $4^n$ and $6^n$.
We know that $\sum \frac{4^n}{6^n}$ converges. Why?
State the comparison test. Is it possible to use the comparison test?
State the limit comparison test. Can we use that?
we went over the problem, the series of (sin(pi/2(2n-1)))/nlnn in class today, and we were asked if it converged. He went over it in class using the alternating series test. I did not really understand it. Could you explain?
ReplyDeleteThe main part is to figure out the behavior of $$\sin\left( \frac{\pi}{2} (2n-1) \right)$$.
Delete$n=1$ gives $1$
$n=2$ gives $-1$
$n=3$ gives $1$
$n=4$ gives $-1$
...
To see this we can simplify the original expression to $\sin\left( \pi n -\frac{\pi}{2} \right)$
Thus for $n=1$ we are starting at the angle $\pi/2$. And every subsequent $n$ adds a factor of $\pi$. Thus, we continue to switch back and forth between the angles $\pi/2$ and $3\pi/2$.
We conclude $$\sin\left( \frac{\pi}{2} (2n-1) \right)=(-1)^{n-1}$$.
Thus we're really working with $$\sum (-1)^{n-1} \frac{1}{n \ln(n)}$$ which is an alternating series, because the terms $b_n = 1/(n\ln(n))$ are positive.
(I assume the series starts at $n>1$, because $\ln(n)\neq 0$.)
We can apply the alternating series test, because the terms are decreasing and $b_n \to 0$ as $n\to \infty$.
(Remark) Here's one way to show the terms are decreasing. To clarify we want to show $b_{n+1}<b_n$ or $1/\left((n+1) \ln (n+1)\right)<1/\left(n\ln(n)\right)$ or $n \ln(n)< (n+1)\ln(n+1)$.
But $0 < n < n+1$ and $0 < \ln(n) < \ln(n+1)$, so we have $n\ln(n) < (n+1)\ln(n+1)$.
Professor Brown said we should know how to approximate for one of the tests. I am doing practice problems, but it would be nice to go over some in section tomorrow to strengthen our understanding! Thanks!
ReplyDeleteHi Tim! After learning numerous tests for series convergence I'm having trouble knowing which test is best for each type of series. If we could do a lot of practice problems with recognition during section that would be great!
ReplyDelete-Erica
I agree! Determining which test fits a particular series is something I find most challenging.
DeleteToday for section 1 I addressed problems 2, 4, 6, 8, 10, 12, and 14 on page 740. In section 2 I addressed problems 2, 4, 6, 8, and 10.
DeleteI will continue going through more evens in class. I encourage read section 11.7 and the 6 examples in that section. Then continue doing the odd problems. Getting help when necessary.
tl;dr:
11.7.2 is done most easily with the root test. The ratio test is also a viable option. "7. If $a_n$ is of the form $(b_n)^n$, then the Root Test may be useful."
I stumbled a bit in the limit comparison test for number 2, but the following would be how.
We wish to compare $$\sum \frac{(2n+1)^n}{n^{2n}}$$ and $$\sum \frac{2^n}{n^n}$$. Then we write
$$\lim_{n\to\infty} \frac{\frac{(2n+1)^n}{n^{2n}}}{\frac{2^n}{n^n}}=\lim_{n\to\infty}\left(1+\frac{1}{2n}\right)^n$$
This limit is of the type $1^\infty$ so we rewrite it as $$\lim \exp(n \cdot \ln (1+\frac{1}{2n}))$$. The limit on the inside of $\exp$ is of the form $\infty \cdot 0$ so we rewrite it as $$\frac{\ln\left(1+\frac{1}{2n}\right)}{1/n}$$. Then apply L'H when taking the limit to get $$\frac{1}{1+\frac{1}{2n}}\cdot \frac{1}{2}$$ which converges to $1/2$ so that our original limit converges to $\exp(1/2)$.
Of course, all this work can be avoided by using the root test.
The idea of the integral test is a little bit weird to me. If it's the best way to determine if a series is converging or diverging, why wasn't this the first thing they showed us when talking about series and why show us now, if we can do it without using it(as we did in 11.2)?
ReplyDeleteI determined several reasons for you. Though there may be more.
DeleteFirst, some of the examples in section 11.2 are concrete and better illustrate the possible behaviors of a series.
Second, the integral test requires that we are able to evaluate or determine the convergence of the function. This isn't always as easy as some of the other tests. It might not even be possible to use the test, as we require the function be non-negative and (ultimately) decreasing.
Third, we are able to evaluate a geometric series explicitly. While the integral test can give us an estimate of the series, it does not give us the precise value.
Fourth, the way the integral test has been introduced, we may be under the lopsided impression that we are to only use the convergence or divergence of an integral to determine the convergence or divergence of a series. However, we can use the theorem to determined whether or not an integral converges by analyzing the corresponding series (provided the hypotheses are satisfied).
If anybody has some other reasons, do share.
With all of these tests for convergence etc. I was getting kind of confused on the order to use when approaching a problem where it is asked if a series converges or diverges. Is there a 'right' order to approaching this problem?
ReplyDeleteMichelle Bohrson
There's no right order, but looking at (as well as learning/memorizing) section 11.7 will definitely help you in organizing your approach to testing the convergence or divergence of a series.
DeleteHi Tim. When we use the Limit Comparison Test is it safe to say we always use the higher degree terms when trying to make our new sequence "b sub n" or are there exceptions.
ReplyDeleteAt the top of my head, I can't think of any exceptions (examples where using the higher degree terms isn't the right choice). We concern ourselves with the limiting behavior of the terms, which is dictated by those higher degree terms.
DeleteI'll leave you with the opinion that using the higher degree terms is a good rule of thumb.
Hi Tim,
ReplyDeleteregarding the homework problem 11.4.20, using the comparison test, I don't know which one should be my a and b for the limit comparison test.
thank you,
Nahyr Lugo-Fagundo
It doesn't matter.
DeleteIf $$\lim \frac{a_n}{b_n} = c$$ for some finite, positive $c$, then $$\lim \frac{b_n}{a_n} = d$$ for some finite, positive $d$. In fact, $$d=\frac{1}{c}$$.
Tim, section today was very helpful. The recognition problems you put on the board and went through were very clear.
ReplyDeleteThanks again!
Erica
I am having trouble with the homework problems for section 11.4 because I do not understand how to pick the series "bn" we should compare the given one "an" to when using the Limit Comparison Test. How or what should we choose to compare our problem to?
ReplyDeleteSome rule of thumbs are to first try looking for a geometric series or power series to which you can compare the original series with. If not, try to pick out the "faster" growing terms and seeing if any of the other convergence or divergence tests apply. Any situation where you can discuss the convergence or divergence of the new series is a "good" candidate for being the series $b_n$. Now try to apply the limit comparison test.
DeleteI feel confident with converging and diverging series, but when there are trig functions in the series I always second guess myself. During next section can we do some problems that involve finding the divergence/convergence of a series with trig functions?
ReplyDeleteThanks!
-Erica
Sure, you'll have to remind me. Though for now, I'd suggest that there are two ways in which trigonometric functions will come into play:
Delete1) They turn the series into an alternating series. For example, when you see
$$\sin(\pi n+ \pi/2)$$ or $$\cos(\pi n)$$. The key here is the $\pi$ that's placed next to the $n$.
2) We don't get a nice pattern from the trigonometric function, but at least we know it's bounded. In which case, we can try to see if the series is absolutely convergent. (Section 11.6)
For example: $$\sum \frac{\sin(n)}{n^2}$$ is absolutely convergent.
3) The limit of the terms does not go to 0, so we conclude the series diverges.
Remark: There may be overlap, that is, situations where you can test for absolute convergence or use alternating series test. Example: $$\sum \frac{\sin (\pi n)}{n^2 }$$. Of course, the benefit of identifying a series as an alternating series is that we can find an estimate of the value of the series.
I'm confused as to how I would find if a series containing a natural log converges or diverges (like 11.7.27).
ReplyDeleteAlso, when a function has a trig function in it where the series isn't truly alternating (since it doesn't go positive, negative, positive, negative), like 11.7.24, can I just say automatically that it diverges? Or is there another test you would recommend?
I'll have to look at the problem, but my intuition is that most problems with the natural log will involve the integral test.
DeleteFor a trig function where the series isn't alternating, you'll want to consider if the series is absolutely convergent. See section 11.6 and my reply to "ericazehnder" directly above.
By the way, thank you for referencing problems along with your questions. If applicable, I will elaborate on my answers when I next have my book.
I'm confused about in the ratio and root test if the third option comes up where the limit is equal to 1 what our next step should be.
ReplyDeleteThanks for your help,
Michelle Bohrson
Section 11.7 pretty much summarizes your options. It even says the ratio and root test will definitely fail for specific type of series. For example, when the terms are quotients of polynomials, the limit is 1. Of course, if the terms are quotients of polynomials, then you should use the limit comparison test combined with the p-test.
DeleteSuch form recognition takes practice. Hopefully you'll get used to it.
Enjoy!
An alternating series is typically not absolutely convergent, correct? For problems like those in 11.6 would we have to test for this if we have found it to be convergent?
ReplyDeleteThanks,
Bailey Richards
An alternating series is not necessarily absolutely convergent. It may not even be convergent.
DeleteFor example:
$$\sum_{n=1}^\infty (-1)^n$$
Hey Tim,
ReplyDeleteFirst, I found the practice tests helpful, thank you.
Here are my posts for the semester:
What is the best type of problem to use the integral test?
For polar problems and finding the area, what is the best strategy to find the limits of integration?
Do you know of the cover up method for partial fractions and can we use it on our test? (as in would we still be showing enough work?)
In series, when do trig functions cause the series to alternate and when do they cause them to converge? I know inverse tan converges to pi/2 and cos(npi) alternates. Doesn't sin(npi) alternate or sinpi?
When you use the alternating series test, are you testing absolute convergence too?
For an alternating series where bn+1 is not less than bn, what should you do?
When finding interval of convergence you sometimes get x-2< 5....in some problems it is ok to move the two to the other side of the inequality, and sometimes it is not?
For trig substitution, in some book examples the limits of integration are changed, but this is not required as long as you put the integral back in terms of x, correct?
In linear diff eqs, an integrating factor is not always necessary, right?
How soon after the final will we know our final grade?
Are you holding office hours?
Thank you for everything this semester.